//Given an array, rotate the array to the right by k steps, where k is non-negat
//ive. 
//
// 
// Example 1: 
//
// 
//Input: nums = [1,2,3,4,5,6,7], k = 3
//Output: [5,6,7,1,2,3,4]
//Explanation:
//rotate 1 steps to the right: [7,1,2,3,4,5,6]
//rotate 2 steps to the right: [6,7,1,2,3,4,5]
//rotate 3 steps to the right: [5,6,7,1,2,3,4]
// 
//
// Example 2: 
//
// 
//Input: nums = [-1,-100,3,99], k = 2
//Output: [3,99,-1,-100]
//Explanation: 
//rotate 1 steps to the right: [99,-1,-100,3]
//rotate 2 steps to the right: [3,99,-1,-100]
// 
//
// 
// Constraints: 
//
// 
// 1 <= nums.length <= 105 
// -231 <= nums[i] <= 231 - 1 
// 0 <= k <= 105 
// 
//
// 
// Follow up: 
//
// 
// Try to come up with as many solutions as you can. There are at least three di
//fferent ways to solve this problem. 
// Could you do it in-place with O(1) extra space? 
// 
// Related Topics 数组 数学 双指针 
// 👍 994 👎 0


package leetcode.editor.cn;

//Java：Rotate Array
class P189RotateArray {
    public static void main(String[] args) {
        Solution solution = new P189RotateArray().new Solution();
        // TO TEST
        solution.rotate(new int[]{1, 2, 3, 4, 5, 6, 7}, 2);
    }

    //leetcode submit region begin(Prohibit modification and deletion)
    class Solution {
        public void rotate(int[] nums, int k) {
            int len = nums.length;
            int rotate = k % len;
            //7,6,5,4,3,2,1

            reverse(0, len - 1, nums);
            reverse(0, rotate - 1, nums);
            reverse(rotate, len - 1, nums);
        }

        public void reverse(int start, int end, int[] arr) {
            while (start < end) {
                swap(start++, end--, arr);
            }
        }

        public void swap(int i, int j, int[] arr) {
            int temp = arr[i];
            arr[i] = arr[j];
            arr[j] = temp;
        }

    }
//leetcode submit region end(Prohibit modification and deletion)

}